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I am studying a bit granular dynamics and I have seen that two spheres of radius $R$ in contact with a contact area of radius $a$ would need an applied force $F$ on this two spheres that is nonlinear in the depth of deformation $\delta$ as it goes as:

$F \sim \delta^{3/2}$

To be honnest, I am not really interested in the full calculation as I am pretty sure I will forget it within two days and plus te full calculation probably would not give me a lot of insight on what is happening.

One way "a la de Gennes" that I have seen consist in relating the stess $\sigma$ to the vertical deformation $\epsilon$ via $\sigma = E \epsilon$.

Then people say that for spheres and if $a \ll R$ then

  • 1) $\epsilon \sim \delta /a$

  • 2) $a \sim \sqrt{\delta R}$

  • $\Rightarrow \:\sigma \sim E\sqrt{\delta/R}$

  • 3) $F = \pi a^2 \sigma \sim \pi R\delta \sqrt{\delta /R} \sim \pi E\sqrt{R}\delta^{3/2}$

This final result is pretty close to the actual one. My point is that I don't understand the physics of the first equation as I am used to $\epsilon = \delta L/L$ which would give me $\epsilon \sim \delta/R$.

Question:

Is there any way to understand physically the fact that $\epsilon \sim \delta/a$?

Thanks very much for any answer.

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On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any different?

No. All of the stress and deformation is local to the contact patch so the size of the overall body makes no difference. Specifically, the deformation only goes to a depth that's proportional to the radius of the contact patch. So the strain, which is inversely proportional to deformation depth, must also be inversely proportional to contact radius.

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thanks for the reply, but for that logic to explain why/how $\epsilon \sim \delta /a $, I would need a somewhat generalised definition of $\epsilon$, because, as I pointed out in my question the usual definition doesn't seem to lead to the same conclusion. –  gatsu Aug 29 at 18:09
1  
Oh, sorry $\epsilon$ is always local percentage compression (not expressed as a percent) so if it's constant across a length, then it will equal $\delta L / L$ but if it varies, you can measure L between any two points to get the average strain between them. So if you measure between two points very close to each other you can get the local strain. For the sphere there is very high strain at the contact patch, but relatively low strain everywhere else so measuring the average strain to the center of the sphere $\delta / R$ would not be proportional to the strain near the surface. –  Rick Aug 30 at 12:45

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